Integral of sin(4x-3)cos(x+5) dx

1. Rewrite the integrand:

   $$\sin{\left(4 x - 3 \right)} \cos{\left(x + 5 \right)} = 8 \sin{\left(5 \right)} \sin^{4}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)} - 8 \sin^{3}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)} - 4 \sin{\left(5 \right)} \sin^{2}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)} + 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 4 \sin{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)} - 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} - 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{5}{\left(x \right)} + 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)} - \sin{\left(3 \right)} \cos{\left(5 \right)} \cos{\left(x \right)}$$

2. Integrate term-by-term:

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 8 \sin{\left(5 \right)} \sin^{4}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)}\, dx = 8 \sin{\left(5 \right)} \cos{\left(3 \right)} \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

      1. Let $u = \sin{\left(x \right)}$.

         Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

         $$\int u^{4}\, du$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         Now substitute $u$ back in:

         $$\frac{\sin^{5}{\left(x \right)}}{5}$$

      So, the result is: $\frac{8 \sin{\left(5 \right)} \sin^{5}{\left(x \right)} \cos{\left(3 \right)}}{5}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- 8 \sin^{3}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \cos{\left(3 \right)} \cos{\left(5 \right)} \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

      2. There are multiple ways to do this integral.

         Method #1

         1. Let $u = \cos{\left(x \right)}$.

            Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$:

            $$\int \left(u^{4} - u^{2}\right)\, du$$

            1. Integrate term-by-term:

               1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                  $$\int u^{4}\, du = \frac{u^{5}}{5}$$

               1. The integral of a constant times a function is the constant times the integral of the function:

                  $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                     $$\int u^{2}\, du = \frac{u^{3}}{3}$$

                  So, the result is: $- \frac{u^{3}}{3}$

               The result is: $\frac{u^{5}}{5} - \frac{u^{3}}{3}$

            Now substitute $u$ back in:

            $$\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$$

         Method #2

         1. Rewrite the integrand:

            $$\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

         2. Integrate term-by-term:

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

               1. Let $u = \cos{\left(x \right)}$.

                  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

                  $$\int u^{4}\, du$$

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

                     1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                        $$\int u^{4}\, du = \frac{u^{5}}{5}$$

                     So, the result is: $- \frac{u^{5}}{5}$

                  Now substitute $u$ back in:

                  $$- \frac{\cos^{5}{\left(x \right)}}{5}$$

               So, the result is: $\frac{\cos^{5}{\left(x \right)}}{5}$

            1. Let $u = \cos{\left(x \right)}$.

               Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

               $$\int u^{2}\, du$$

               1. The integral of a constant times a function is the constant times the integral of the function:

                  $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                     $$\int u^{2}\, du = \frac{u^{3}}{3}$$

                  So, the result is: $- \frac{u^{3}}{3}$

               Now substitute $u$ back in:

               $$- \frac{\cos^{3}{\left(x \right)}}{3}$$

            The result is: $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$

         Method #3

         1. Rewrite the integrand:

            $$\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

         2. Integrate term-by-term:

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

               1. Let $u = \cos{\left(x \right)}$.

                  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

                  $$\int u^{4}\, du$$

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

                     1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                        $$\int u^{4}\, du = \frac{u^{5}}{5}$$

                     So, the result is: $- \frac{u^{5}}{5}$

                  Now substitute $u$ back in:

                  $$- \frac{\cos^{5}{\left(x \right)}}{5}$$

               So, the result is: $\frac{\cos^{5}{\left(x \right)}}{5}$

            1. Let $u = \cos{\left(x \right)}$.

               Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

               $$\int u^{2}\, du$$

               1. The integral of a constant times a function is the constant times the integral of the function:

                  $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                     $$\int u^{2}\, du = \frac{u^{3}}{3}$$

                  So, the result is: $- \frac{u^{3}}{3}$

               Now substitute $u$ back in:

               $$- \frac{\cos^{3}{\left(x \right)}}{3}$$

            The result is: $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$

      So, the result is: $- 8 \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) \cos{\left(3 \right)} \cos{\left(5 \right)}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- 4 \sin{\left(5 \right)} \sin^{2}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)}\right)\, dx = - 4 \sin{\left(5 \right)} \cos{\left(3 \right)} \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

      1. Let $u = \sin{\left(x \right)}$.

         Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

         $$\int u^{2}\, du$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

         Now substitute $u$ back in:

         $$\frac{\sin^{3}{\left(x \right)}}{3}$$

      So, the result is: $- \frac{4 \sin{\left(5 \right)} \sin^{3}{\left(x \right)} \cos{\left(3 \right)}}{3}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

      1. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

         $$\int u^{4}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u^{4}\, du = \frac{u^{5}}{5}$$

            So, the result is: $- \frac{u^{5}}{5}$

         Now substitute $u$ back in:

         $$- \frac{\cos^{5}{\left(x \right)}}{5}$$

      So, the result is: $- \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{5}{\left(x \right)}}{5}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 4 \sin{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)}\, dx = 4 \cos{\left(3 \right)} \cos{\left(5 \right)} \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

      1. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

         $$\int u^{2}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            So, the result is: $- \frac{u^{3}}{3}$

         Now substitute $u$ back in:

         $$- \frac{\cos^{3}{\left(x \right)}}{3}$$

      So, the result is: $- \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}}{3}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

      1. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

         $$\int u^{2}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            So, the result is: $- \frac{u^{3}}{3}$

         Now substitute $u$ back in:

         $$- \frac{\cos^{3}{\left(x \right)}}{3}$$

      So, the result is: $\frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{3}{\left(x \right)}}{3}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)}\, dx = \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)}\, dx$$

      1. The integral of sine is negative cosine:

         $$\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$$

      So, the result is: $- \sin{\left(3 \right)} \sin{\left(5 \right)} \cos{\left(x \right)}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{5}{\left(x \right)}\right)\, dx = - 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos^{5}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}$$

      2. Rewrite the integrand:

         $$\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$$

      3. Integrate term-by-term:

         1. Let $u = \sin{\left(x \right)}$.

            Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

            $$\int u^{4}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u^{4}\, du = \frac{u^{5}}{5}$$

            Now substitute $u$ back in:

            $$\frac{\sin^{5}{\left(x \right)}}{5}$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

            1. Let $u = \sin{\left(x \right)}$.

               Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

               $$\int u^{2}\, du$$

               1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                  $$\int u^{2}\, du = \frac{u^{3}}{3}$$

               Now substitute $u$ back in:

               $$\frac{\sin^{3}{\left(x \right)}}{3}$$

            So, the result is: $- \frac{2 \sin^{3}{\left(x \right)}}{3}$

         1. The integral of cosine is sine:

            $$\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$$

         The result is: $\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$

      So, the result is: $- 8 \left(\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}\, dx = 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos^{3}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$$

      2. Let $u = \sin{\left(x \right)}$.

         Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

         $$\int \left(1 - u^{2}\right)\, du$$

         1. Integrate term-by-term:

            1. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

               1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                  $$\int u^{2}\, du = \frac{u^{3}}{3}$$

               So, the result is: $- \frac{u^{3}}{3}$

            The result is: $- \frac{u^{3}}{3} + u$

         Now substitute $u$ back in:

         $$- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$$

      So, the result is: $8 \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- \sin{\left(3 \right)} \cos{\left(5 \right)} \cos{\left(x \right)}\right)\, dx = - \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos{\left(x \right)}\, dx$$

      1. The integral of cosine is sine:

         $$\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$$

      So, the result is: $- \sin{\left(3 \right)} \sin{\left(x \right)} \cos{\left(5 \right)}$

   The result is: $8 \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)} - 8 \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) \cos{\left(3 \right)} \cos{\left(5 \right)} - 8 \left(\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)} + \frac{8 \sin{\left(5 \right)} \sin^{5}{\left(x \right)} \cos{\left(3 \right)}}{5} - \frac{4 \sin{\left(5 \right)} \sin^{3}{\left(x \right)} \cos{\left(3 \right)}}{3} - \sin{\left(3 \right)} \sin{\left(x \right)} \cos{\left(5 \right)} - \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{5}{\left(x \right)}}{5} + \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{3}{\left(x \right)}}{3} - \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}}{3} - \sin{\left(3 \right)} \sin{\left(5 \right)} \cos{\left(x \right)}$

3. Now simplify:

   $$- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}$$

4. Add the constant of integration:

   $$- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}+ \mathrm{constant}$$